What can be said about $\gcd(N/q^{\alpha},\sigma(N/q^{\alpha}))$ where $N$ is an odd perfect number and $q^{\alpha} \parallel N$?

Question

What can be said about the quantity $$\gcd(N/q^{\alpha},\sigma(N/q^{\alpha}))$$ where $N$ is an odd perfect number and $q^{\alpha} \parallel N$? In particular, can one prove that it is always greater than one?

Note that, when $q = p$ is the special prime of $N$, then since we can express $N$ as $$N = p^k m^2$$ (where $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$), we obtain $$\gcd(N/q^{\alpha},\sigma(N/q^{\alpha})) = \gcd(m^2,\sigma(m^2)) = \sigma(m^2)/p^k = \sigma(N/q^{\alpha})/q^{\alpha},$$ which is always greater than one.

The situation is much more complicated when $q \neq p$.

Answer 1

From the comments:

$$\gcd(N/q^\alpha,\sigma(N/q^\alpha)) = \begin{cases} \dfrac{N}{q^\alpha \sigma(q^\alpha)}, & \alpha \equiv 0 \pmod 2, \\[8pt] \dfrac{2N}{q^\alpha \sigma(q^\alpha)}, & \alpha \equiv 1 \pmod 2. \end{cases}$$

Since it is known that the index $$\sigma(N/q^{\alpha})/q^{\alpha} \geq 3$$ and since $$\sigma(q^{\alpha})\sigma(N/q^{\alpha})=\sigma(N)=2N,$$ because $N$ is perfect and $\gcd(q^{\alpha},N/q^{\alpha})=1$, then we obtain $$\sigma(N/q^{\alpha})/q^{\alpha} = \frac{2N}{q^{\alpha} \sigma(q^{\alpha})}.$$

It follows that

$$\gcd(N/q^{\alpha},\sigma(N/q^{\alpha})) \geq \dfrac{3}{2} > 1,$$

which proves the original claim.